$\overline{AC} = 10$ $\overline{BC} = {?}$ A C B 10 ? $ \sin( \angle ABC ) = \frac{10\sqrt{109} }{109}, \cos( \angle ABC ) = \frac{3\sqrt{109} }{109}, \tan( \angle ABC ) = \dfrac{10}{3}$
Explanation: $\overline{AC}$ is the opposite to $\angle ABC$ $\overline{BC}$ is adjacent to $\angle ABC$ SOH CAH TOA We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA) $ \tan( \angle ABC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{AC}}{\overline{BC}}= \frac{10}{\overline{BC}} $ $ \overline{BC}=\frac{10}{\tan( \angle ABC )} = \frac{10}{\dfrac{10}{3}} = 3$